\newproblem{lay:1_1_16}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.16}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Clara Susana Rey Abad, Nov. 4, 2013} \\}{}

  % Problem statement
  Determine whether the following system is consistent (do not fully solve the system).
	\begin{center}
		$\begin{array}{rrrrcr}
			2x_1&       &       & -4x_4 & = & -10 \\
			    & +3x_2 & +3x_3 &       & = &  0 \\
		      &       & + x_3 & +4x_4 & = & -1 \\
		 -3x_1& +2x_2 & +3x_3 & + x_4 & = &  5
		\end{array}$
	\end{center}
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrrr|r}
		   2 &  0 &  0 & -4 & -10 \\
		   0 &  3 &  3 &  0 &  0 \\
		   0 &  0 &  1 &  4 & -1 \\
		  -3 &  2 &  3 &  1 &  5 \\
		\end{array}\right)$
	\end{center}
	Now, we apply row operations to solve it
	\begin{center}
		\begin{tabular}{cc}
			 $\begin{array}{c}\mathbf{r}_1\leftarrow \mathbf{r}_1\div 2\end{array}$ &
				$\left(\begin{array}{rrrr|r}
					 1 &  0 &  0 & -2 & -5 \\
					 0 &  3 &  3 &  0 &  0 \\
					 0 &  0 &  1 &  4 & -1 \\
					-3 &  2 &  3 &  1 &  5 \\
				\end{array}\right)$ \\
			 $\mathbf{r}_2\leftarrow \mathbf{r}_2\div 3$ &
				$\left(\begin{array}{rrrr|r}
					 1 &  0 & 0 & -2 & -5 \\
					 0 &  1 & 1 &  0 &  0 \\
					 0 &  0 & 1 &  4 & -1 \\
					-3 &  2 & 3 &  1 &  5 \\
				\end{array}\right)$ \\
			 $\mathbf{r}_4\leftarrow \mathbf{r}_4+3\mathbf{r}_1$ &
				$\left(\begin{array}{rrrr|r}
					 1 &  0 & 0 & -2 & -5 \\
					 0 &  1 & 1 &  0 &  0 \\
					 0 &  0 & 1 &  4 & -1 \\
					 0 &  2 & 3 & -5 & -10 \\
				\end{array}\right)$ \\
			$\mathbf{r}_4\leftarrow \mathbf{r}_4-2\mathbf{r}_2$ &
			 $\left(\begin{array}{rrrr|r}
			     1 &  0 & 0 & -2 & -5 \\
					 0 &  1 & 1 &  0 &  0 \\
					 0 &  0 & 1 &  4 & -1 \\
					 0 &  0 & 1 & -5 & -10 \\
				\end{array}\right)$ \\
			$\mathbf{r}_4\leftarrow \mathbf{r}_4-\mathbf{r}_3$ &
			 $\left(\begin{array}{rrrr|r}
			     1 & 0  & 0 & -2 & -5 \\
					 0 & 1  & 1 &  0 &  0 \\
					 0 & 0  & 1 &  4 & -1 \\
					 0 & 0  & 0 & -9 & -9 \\
				\end{array}\right)$ \\
		\end{tabular}
	\end{center}
	
  The system is compatible since there are four equations and four leading entries.
}
\useproblem{lay:1_1_16}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
